Finn's Maths Help!

Circle Theorems

Here's a list of every circle theorem you'll need for GCSE Maths :)

Parts of a circle:

A chord is a line going across a circle. It creates a segment.

The perpendicular bisector of any chord can be extended to create a radius/diameter.
This dotted line is the perpendicular bisector. It's a straight line that passes straight through the middle of the chord, at a right angle.
It is therefore also the diameter of the circle.

A tangent is a line that touches a circle at one point only. The angle between the tangent and the radius is always 90�.
This line at the bottom is the tangent.

Tangents drawn to a circle from a point outside the circle are equal in length. AC = BC, in this example.
So we have the point C. Lines CA and CB are tangents, and so because they both originate from the point C outside the circle, they are the same length.

The angle at the centre of a circle is twice the angle at the circumference, when both angles are subtended by the same arc.
Help! What does subtended mean?!
Well, the angle C is created by the two chords AC and BC. So, we say the angle C is subtended by the arc (curve) between A and B.
As you can see, the angle at the centre of the circle is subtended by the same arc, and is therefore twice as large as the angle C.

This theorem still works if the angle is off-centre. As long as the two angles are subtended by the same arc, the centre angle will be twice as large as the circumference angle.

The angle in a semicircle is a right angle. Any triangle joining both ends of the diameter to a point on the circumference is a right-angled triangle.
In this example, AB is the diameter of the circle, and C is a point on the circumference.
The ends of the diameter, A and B, join up to C, and create a right-angled triangle.

Angles subtended by the arc (or chord) AB in the same segment are equal.
The two angles marked here are both subtended by the same arc and are in the same segment. This means that they are equal in size.

Opposite angles of a cyclic quadrilateral add to 180�.
A cyclic quadrilateral is any quadrilateral drawn inside a circle. Every corner of the quadrilateral must touch the circumference of the circle, like in this picture.
In this example, a + b = 180�, and x + y = 180�.

Alternate Segment Theorem

The angle between the tangent and the chord AB is equal to the angle in the alternate segment.
In this circle, the chord AB creates 2 segments (it cuts the circle into 2 bits).
The tangent line, AT, creates the angle x with this chord, in one of the segments.
In the other segment, an angle has been created at point C from the chord AB. This angle is also x, because it is in the alternate segment.

Equations of Circles

A circle is basically a line.
Finding the equation of a circle just means finding the equation of a circular line.
To find the equation of a circle, you have to draw a little right-angled triangle.

The hypotenuse of the triangle will be the radius (remember that the radius can be drawn from the centre to any point on the circumference). In this circle, the radius is 4. So the hypotenuse is 4.
The other two sides of the triangle will create the right-angle, so it's easiest to draw this on a grid. We can call these sides x and y.

The equation of the circle is simple once you've done this: it's x� + y� = r�. r is the radius.
So, in this example, it's x� + y� = 16.

Equations of Tangents

Here are the steps for finding the equation of a tangent on a grid with coordinates.

Draw a triangle, like we did for finding the equation of a circle.
Find the lengths of the vertical and horizontal sides of the triangle. In this example, they are 3 and 4. (You find these by looking at the coordinates and counting).
Remember that the tangent is perpendicular to the radius; they make a right angle.
Find the gradient of the radius OA (the vertical divided by the horizontal side lengths). In this example, the gradient of the radius is 4/3.
The gradient of the tangent is the negative reciprocal of the gradient of the radius. In this example, the gradient of the tangent is -3/4.
The tangent is a straight line, so its equation is y = mx + c. m is the gradient, which is -3/4 in this example, so we can substitute that in to make y = -3/4x + c.
Now, substitute in the x and y values from point A. In this example, y = 4 and x = 3. Simplify this equation to get 4 = -9/4 + c.

Solve this equation to get the value of c, which is the y-intercept of the tangent. In this example, it is 6 1/4.
Create the complete equation using the gradient and y-intercept of your tangent!
The equation for this tangent is y = -3/4x + 6 1/4.